Science desk
< July 5	<< Jun \| July \| Aug >>	July 7 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

July 6

Voltage to Current converter (2A)

Is there a way to simply generate a 0-2 A current source proportionally to a 0-5 V or 0-10 V control voltage signal. I would like to replicate the functionality of this device. --Jcmaco 00:46, 6 July 2007 (UTC)[reply]

That would be an amplifier. Do you intend to design the circuit, or are you hoping to buy a commercially available one? In either case, you will need more thorough specification of requirements (what load will the 2 amps be driven into? How much variation can you tolerate?) 2 amps may be a lot of juice, so be careful if you try to build it yourself. Nimur 00:51, 6 July 2007 (UTC)[reply]

The load is variable (4-8 ohm). I intend to design and built the circuit myself since the available commercial solutions are too expensive. Could a high current Op-Amp (like the OPA548) be used if I were to control the current limit? --Jcmaco 01:02, 6 July 2007 (UTC)[reply]

See current source to start with--Tugjob 01:29, 6 July 2007 (UTC)[reply]

Yes, you want a voltage to current converter. see Voltage-to-current converter and this for some ideas. --Duk 05:54, 6 July 2007 (UTC)[reply]

Spherical mirror

After seeing the result of what happens when two mirrors face each other, I wondered what things would look like inside a spherical mirror. Then I began to wonder about the physics of such a situation. Imagine you are inside a spherical object or room where the inner surface is a highly reflective mirror. If you turn on a flashlight for a brief moment and turn it off again, how long would you be able to see, since most of the energy is constantly being reflected? Some of the energy would be absorbed by the mirrored surface, some by your body, some would impact the eyes and be absorbed by the retina, etc. but surely most of it would be reflected many times before being absorbed. Or would the shape of the spherical mirror focus the light on the center of the sphere? Suppose you turned the flashlight on and left it on. Would the trapped light significantly raise the temperature inside the sphere? 152.16.59.190 07:23, 6 July 2007 (UTC)[reply]

Don't forget that conservation of energy still applies, so it won't get much warmer, at least not quickly. About focusing: See curved mirror - a perfectly focusing mirror is actually paraboloid, not spherical. A small segment of your sphere, at which the light from the flashlight hits the surface nearly at an angle of 90 degrees, can be viewed as an approximation to a paraboloid, but then the distance from the surface to the focal point is half the radius of the sphere. Icek 09:01, 6 July 2007 (UTC)[reply]

Even the very best telescope mirrors absorb 5–10% of the energy (each time the light is reflected), so it wouldn't take long for almost all the energy to be absorbed.--Shantavira|^{feed me} 09:29, 6 July 2007 (UTC)[reply]

Even if you ignore everything inside the mirror, the light won't last for long. Say the mirror was 99.9999% reflecting (that's about the absolute best possible mirror you can make, but in reality such a mirror would only reflect that well over a narrow band of wavelengths and reflection angles - we'll ignore that for now). So the mirror reflectivity is R=0.999999. After N bounces of the light from the mirror, the intensity of the remaining light is I = I0 * R^N, where I0 is the light's starting intensity. Now say the sphere is such a size that light travels an average of 1 metre between each bounce. So the time taken to make N bounces is just N/speed of light. Now work out those numbers; after 50 milliseconds, the light has made 150 million bounces, and the intensity of the light is 1/(30 millionth) of what it started at. So only 50 milliseconds after you turn off the light, the remaining light bouncing inside the sphere is essentially undetectable. --Bob Mellish 09:36, 6 July 2007 (UTC)[reply]

I don't think that a flashlight left on would increase the temperature significantly. In every reflection, only a part of the energy is being absorbed (and the remaining reflected). At the end of all reflections, the mirror would have absorbed all the energy. This energy absorbed would be equal to the energy absorbed by a black body (without any reflection). I don't expect a (non reflecting) black spherical mirror to become very hot due to a torch light left inside it and hence a reflecting mirror would be even less hotter -- WikiCheng | Talk 12:30, 6 July 2007 (UTC)[reply]

"Black mirror". What? Capuchin 13:07, 6 July 2007 (UTC)[reply]

No matter what - all of the heat energy that's created during this process came from a couple of little batteries inside the flashlight. The total amount simply cannot be more than that. Even if the outside of the sphere is exceedingly well insulated, the amount of heat gained won't be all that much. The situation is similar to the parallel mirrors case in that the light would bounce around forever if it were not for the fact that no mirror is perfect - and anything you place between the mirrors/within the sphere such as yourself and your flashlight - will inevitably absorb light themselves. However, small the imperfection is, it'll eventually end up absorbing all of the light and turning it into heat. SteveBaker 13:24, 6 July 2007 (UTC)[reply]

I have measured a flashlight with 2 D alkaline cells to produce and consume .35 amps at 3 volts, which is 1.05 watts, or about 1 Joule/second. Now take the mass and specific heat of the air and any person inside the sphere, plus the sphere itself, and figure how fast it would heat up, even if it were insulated and no heat left it. Compare that to the heat evolved from the observer, which might be 60 to 80 watts. The flashlight would be a negligible addition. Edison 14:20, 6 July 2007 (UTC)[reply]

... similar to the parallel mirrors case in that the light would bounce around forever ... -- Actually, it's my understanding that light would fall to the ground in the perfectly-reflecting parallel mirrors case. [1] --TotoBaggins 14:32, 6 July 2007 (UTC)[reply]

Hah! I guess that depends on how you define parallel. If the mirrors were aligned perpendicular to spacelike geodesics, then sure, the light would fall. But if they were aligned perpendicular to null geodesics, the light would go back and forth along the same path, because light always follows null geodesics. One could argue that that's a better definition of parallel. In practice, the only way to align the mirrors that accurately is by interferometry... and it's obvious what definition that leads to. —Keenan Pepper 18:05, 6 July 2007 (UTC)[reply]

Missing diffraction spikes

In the picture at right exhibiting diffraction spikes, the spikes seem to be proportional to the brightness of the star. However, in today's Astronomy Picture of the Day (see here), the effect seems to be either off or on, with nothing in between, and not depending strictly on brightness. Explanations I can think of are 1) the image was composed of shots from different telescopes, or 2) the spikes that are there are photoshopped in for effect. Is there some more compelling explanation? Thanks. --TotoBaggins 16:28, 6 July 2007 (UTC)[reply]

I have mentioned before that I believe many of these astronomical photographs use digital effects for artistic purposes, and that it can blur the distinction between "scientific data" from "art." I cannot think of a good reason why those diffraction spikes should only appear selectively, as you mention. Nimur 19:09, 6 July 2007 (UTC)[reply]

In fact, the source of that image specifically states:

“

Separate black and white exposures through clear, red, green and blue filters are digitally combined and stretched, using Adobe Photoshop and other image processing software, to create full color pictures.
--Cosmotography

”

.

It is difficult to know "how much" post-processing is used. Nimur 19:11, 6 July 2007 (UTC)[reply]

I wrote to the photographer, and he replied with a very helpful and friendly response which read in part:

"Diffraction spikes are proportional to the brilliance of the star that causes them. Only a handful of stars had sufficient brightness in this particular image.

Interestingly, the diameter of stars in astronomical images are usually perceived as a metaphor for their relative brightness to one another. In the case of this photograph, some of the brighter stars were digitally reduced (their were diameters decreased) so that would not appear distracting. You can spot these fairly easily- they are the ones with diffraction spikes. Notice that they also have a colorful halo- those halos give a good indication of their original size in the raw data. I purposely left the halos surrounding each star (and their diffraction spikes) as a way of showing the true size of the star and as an substitute metaphor for brightness.

This is a fairly common approach used in astronomical photography, by the way...

Significantly, nothing was artificially added to this image- the diffraction spikes were not drawn onto the picture."

So I guess the answer is not that spikes were added to a few of the stars, but that they were subtracted from a bunch of them. Thanks for the replies, Nimur. --TotoBaggins 20:38, 6 July 2007 (UTC)[reply]

Well, no, spikes were neither added nor subtracted. The brightest stars were dimmed. Diffraction spikes are indeed "proportional to the brightness of the star" as you wrote initially, and in this image noticeable spikes occurred only at the very brightest stars. Because the apparent brightness of these stars was artificially dimmed, the expected proportionality between star brightness and spike brightness was thrown off. Thus you were right to sense that something was strange. --mglg_(talk) 23:53, 6 July 2007 (UTC)[reply]

And if you look at the APOD photo again, you will see that the diffraction spikes on the two bright blue stars to the right of the galaxy are longer than the others, although not by much. --Anonymous, July 6, 2007, 21:45 (UTC).

Nasty Bread

Can potentially deadly types of mold (or other gross stuff) grow on artisan bread if it's left out for a wicked long time?

This source (and my mycologist sister-in-law) say that it is unsafe to eat moldy bread. "Deadly" is relative, of course, as people have survived arsenic and died from peanut butter. The bread being "artisanal" is unlikely to have an effect, as mold doesn't know how lovingly and authentically and with how much integrity its delicious substrate was prepared. --TotoBaggins 20:31, 6 July 2007 (UTC)[reply]

It's generally not safe to eat anything moldy. "Deadly" is definitely relative. People have died from water. Bart133 ^{(t) (c)} 22:02, 6 July 2007 (UTC) (Link changed --Anon, July 6, 23:00 (UTC)).[reply]

Artisan bread wouldn't contain anywhere near the amount of preservatives found in, say, brand-name packaged white bread... would that make a difference in the amount and number of types of organisms able to grow on the bread.

Killroy filter

"Part of a band pass filter" according to Pynchon

I'm no electrical engineer, but isn't this a band-stop filter? —Keenan Pepper 18:11, 6 July 2007 (UTC)[reply]

Correct. To be painfully precise, it is a parallel resonant band-stop filter. -- Kainaw^(what?) 18:23, 6 July 2007 (UTC)[reply]

It depends where you take the output, doesn't it? Nimur 19:15, 6 July 2007 (UTC)[reply]

I'm assuming that the missing part of the circuit (the left/right ends that go off the edges) loop back and connect to some sort of source. See http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/AC/02127.png for an example. -- Kainaw^(what?) 19:18, 6 July 2007 (UTC)[reply]

As shown here

High Pass Filter
Low Pass Filter

Identical components, in the same topology, function as either a pass- or a stop- filter depending on where the input and output are placed. Nimur 19:21, 6 July 2007 (UTC)[reply]

The circuits you have shown are not "parallel resonant" circuits like the one asked about (with the inductor and capacitor in parallel). To turn this into a "pass" filter, the load resister is also put in parallel (see http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/AC/02125.png for an example of a parallel resonant band-pass filter). I must make a disclaimer that I only work with common circuits. I'm sure someone knows how to get these circuits to do things they are not intended to do. However, if you are taking an AC electronics test and they show you an RLC circuit with the cap/inductor in parallel and ask if it is a stop or pass filter. If the load resister is in series, it is stop. If the load resister is in parallel, it is pass. -- Kainaw^(what?) 23:26, 6 July 2007 (UTC)[reply]

I thought someone was having a little fun. Look in the trivia section of Kilroy was here, which was the first thing I thought of when I saw this. Lanfear's Bane

It's Chad! DuncanHill 10:27, 7 July 2007 (UTC)[reply]

Oddly - it's Chad in the UK, Kilroy in the USA. Having said that - when I lived in UK - I saw plenty of Chad's adorned with the words "Kilroy woz 'ere" - so it is perhaps not that clear-cut! SteveBaker 14:01, 7 July 2007 (UTC)[reply]

Enthaply Change

how do i calculate the mean molar bond entalpy of Br-F bonds in Bromine Pentaflouride.

I have th following figures: molar bond enthalpy F-F-159, Br-Br 193

the overall enthalpy change is -429

many thanks

ΔH = ΣBonds broken - ΣBonds formed. So, you have the equation 1/2Br₂ + 5/2F₂ --> BrF₅.

Figure out how many bonds of each there are, and use algebra to find out what you want. 74.102.89.241 22:05, 6 July 2007 (UTC)[reply]

Shaking arm

Why is that when you flex your arm, or any other body part, it begins to shake, even if it is at rest on a table? Imaninjapirate^{talk to me} 21:28, 6 July 2007 (UTC)[reply]

See Muscle contraction. A brief perusal of the article leads me to believe it may have something to do with frequency summation (apparently only approx 30% of the fibers are firing at any given time), but muscular contractions are complex processes so I am sure there is more to it than that. Is there a doctor in the house? 161.222.160.8 22:35, 6 July 2007 (UTC)[reply]

It is not clear what you mean by "flex your arm". You may mean that you are activating both flexors and extensors at the same time and trying consciously to hold the limb in a particular position. This creates a feedback problem for the muscles and nerves, a problem that is not solved efficiently without practice. --JWSchmidt 04:15, 7 July 2007 (UTC)[reply]

BFS?